Question

is there any way u know this​

Answer 1

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AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the center of the circle.

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• Let AB and BC be equal chords of circle with center O.
• Draw the angle bisector AD of ∠BAC.

Join BC, meeting AD at M.

In triangle DAM and CAM:

AB = BC (given that the chords are equal)

∠BAM = ∠CAM (∵ AD is angle bisector of A)

AM = AM (common side)

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∴ ΔABD = ΔACD (by SAS congruence rule)

∴ BM = CM and ∠AMB = ∠AMC = x (say) (by CPCT) ………(1)

But ∠AMB + ∠AMC = 180° ………(2)

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From equation (1) and (2), we have:

x + x = 180°

⇒ 2x = 1280°

⇒ x = 90°

⇒ ∠AMB = ∠AMC = 90°

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Therefore, AD is the perpendicular bisector of chord BC, but the perpendicular bisector of any chord always passes from the center of the circle.

∴ AD passes through the center O of the circle.

Thus, O lies on the angle bisector of the angle BAC.

Hence, proved.

Answer 2

Answer:

yes

Step-by-step explanation:

AB=AC

therefore ABC is an isoceles triangle.....and if perpendicular is drawn to BC then we can prove

In ∆AOC and ∆AOB

AO=AO (common)

AB=AC(given)

Angle AOC=Angle AOB(=90°)

Hence both triangles are congruent

and by C.P.C.T.C OB=OC

Which means O is centre of circle